If $x=3 \tan \mathrm{t}$ and $y=3 \sec \mathrm{t}$, then the value of $\frac{\mathrm{d}^2 y}{\mathrm{~d}…

If $x=3 \tan \mathrm{t}$ and $y=3 \sec \mathrm{t}$, then the value of $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}$ at $\mathrm{t}=\frac{\pi}{4}$ is
  1. $\frac{-1}{6 \sqrt{2}}$
  2. $\frac{1}{6 \sqrt{2}}$
  3. $\frac{1}{3 \sqrt{2}}$
  4. $\frac{3}{2 \sqrt{2}}$

Solution

$\begin{aligned} & x=3 \tan \mathrm{t} \\ & \therefore \quad \frac{\mathrm{d} x}{\mathrm{dt}}=3 \sec ^2 \mathrm{t} \\ & y=3 \sec \mathrm{t} \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{dt}}=3 \sec \mathrm{t} \tan \mathrm{t} \\ & \text { Now, } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{3 \sec \mathrm{tan} t}{3 \sec ^2 \mathrm{t}}=\sin \mathrm{t} \\ & \therefore \quad \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\mathrm{d}}{\mathrm{dt}}(\sin \mathrm{t}) \cdot \frac{\mathrm{dt}}{\mathrm{d} x} \\ & =\cos t \times \frac{1}{3 \sec ^2 t} \\ & \therefore \quad \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\cos ^3 \mathrm{t}}{3} \\ & \therefore \quad\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)_{\left(\mathrm{t}=\frac{\pi}{4}\right)}=\frac{\left(\cos \frac{\pi}{4}\right)^3}{3}=\frac{1}{6 \sqrt{2}} \\ & \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 2)

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